Are there
for any $p<q$, $p,q\in P$ we have $f_q(x)=f_p(x)$ for each $x\in [0,1]$ with $f_p(x)\neq 0$?
On
You are asking in particular, that for some $p$ in your order, $f_q(x) = f_p(x)$ for all $q > p$ whenever $x \neq 0$. This can be written in a more compact and suggestive notation : $$ p < q \qquad \Longleftrightarrow \qquad \mathrm{supp}(f_q) \subseteq \mathrm{supp}(f_p) \quad \text{ and } \quad f_q(x) = f_p(x) \quad \forall x \notin \mathrm{supp}(f_p). $$ In another words, the linear order on $P$ moves naturally to a new linear partial order $\le'$ on the elements of $p$ defined by $p \,\,\, \le' \, q$ if $\mathrm{supp}(f_q) \subseteq \mathrm{supp}(f_p)$. This linear partial order is more "relaxed" because we might have $p < q$ and $\mathrm{supp}(f_q) = \mathrm{supp}(f_p)$, but if this is not the case then $p <' q$ and $p \le' q$ are equivalent. Therefore, non-triviality lies where the induced linear order becomes smaller.
What I am saying is a little off mathematically speaking, because partial orders require $a \le b$ and $b \le a$ means $a = b$, but we can always get this statement by "putting together" in an equivalence class all the $a$'s and $b$'s that satisfy this property with $a \neq b$. Our order we had before was linear, so there won't be any trouble there. I think everyone gets the idea so I won't go any deeper in that direction.
As yoyo points out in his comment, it is easy to get this linear order $\le'$ by letting the set $\{ f_p \, | \, p \in P \}$ take only countably distinct values (by a value, I mean a function from $[0,1]$ to $\mathbb C$), because it would just mean that letting $\{ O_n\}$ be a countable sequence of disjoint open sets in $[0,1]$ and $\{ x_n \}$ a strictly increasing sequence in $P$, we could have functions $f_n$ that are non-zero only in the first $n$ sets $O_i$ and zero everywhere else (plus being continuous, equiboundedness can be acheived quite easily too). By letting $f_p = f_n$ for all $p$ such that $x_n \le p < x_{n+1}$, we could get such a family of functions. Explicit constructions of such families are not interesting at this point, in my opinion.
The interesting part is if there exists a family of uncountably many distinct functions arising from the set $\{f_p \, | \, p \in P\}$.
Let's prove that there is no such family. First important fact : any open set in the real line is written uniquely as a countable union of disjoint open intervals.
Suppose that we let $p \in P$ and consider the "uncountable sequence" $Q = \{ q \in P \, | \, q > p \}$, for which the functions $f_q$ are all distinct and satisfy the properties you require.
For each $q \in Q$, there exists, by hypothesis, at least one $\overline x$ such that $f_q(\overline x) \neq f_r( \overline x)$ for all $r \in Q$ with $r < q$. Take $\overline q$ to be the largest open interval containing $\overline x$ where $f_q(x) \neq 0$ for all $x$ in this interval. By construction, on this interval, $f_r(x) = 0$ (use a continuity argument to complete this part of the proof). This means that $r < q$ implies that $\overline r$ and $\overline q$ are disjoint.
This gives us the set $$ A = \bigcup_{q \in Q} \overline q $$ which is open, and we've just written it as an uncountable union of disjoint open intervals. Such a set cannot exist over the real line, so we obtain a contradiction.
EDIT : As my set theory is a little rusty, S.Serva noticed that I assumed something that was not written ; that there exists $p \in Q$ such that $\{ q \ge p \}$ is uncountable. Well, this assumption is not quite necessary : one of the sets $\{ q \ge p \}$ and $\{q \le p \}$ must be uncountable, for if they were both, $P$ would be countable. If the first one is uncountable my proof is fine. If the first one is not, then $\{ q \le p \}$ is uncountable and we can do the same argument but "upside-down". Here's a sketch of how things go.
Remember we assume there is an uncountable family of distinct functions $f_p$ satisfying the above. Therefore, only one of those functions can have empty support. But if one such function $f_q$ had empty support, then for some $p < q$, $f_q = f_p$ for all $x \in \mathrm{supp}(f_q)$ and by continuity of $f_q$ and $f_p$ we would get a zero of $f_p$ somewhere. Therefore, $\mathrm{supp}(f_q) \neq \varnothing$ for all $q$. The set $$ \bigcap_{p \in P} \mathrm{supp}(f_p) $$ is an intersection of a decreasing sequence of non-empty compact sets (decreasing with respect to the inclusion order), thus is non-empty. Instead of before where I started with the empty set and "added" intervals to form $$ \bigcup_{q \in Q} \overline q, $$ we are going to start with the entire open set, i.e. $$ [0,1] \cap \left( \bigcap_{p \in P} \mathrm{supp}(f_p) \right)^c $$ and show that we can "subtract" uncountably many disjoint open intervals from it. Consider the "uncountable sequence $Q = \{ q < p \}$ for some $p$ and for representatives $q$ of distinct functions in $P$. Therefore, the supports $\mathrm{supp}(f_q)$ are distinct and included in each other (with respect to the order on $P$). For the same reasons as in the preceding case, for each $q \in Q$ we are given an open interval $\overline q$ where $q' < q$ implies that $\mathrm{supp}(f_{q'}) \cap \overline q = \varnothing$. The same ideas finish the job.
Hope that helps,
On
I don't understand why the following (for instance) doesn't answer the question:
Take $P:=[0,1]$ and $f_p(x)=0$ for all $p$ and all $x$ in $P$.
In case the question is edited, here is the current version:
Are there
an uncountable linear order $(P, \leq)$ and
a family of continuous complex-valued functions $\{f_p\colon p\in P\}$ defined on the interval $[0,1]$ such that
for any $p<q$, $p,q\in P$ we have $f_q(x)=f_p(x)$ for each $x\in [0,1]$ with $f_p(x)\neq 0$?
On
I couldn't find a comment button but in the first solution above the choice of $I(p)$, $a_p$ and $L(p)$ (or $R(p)$) does not take the other functions into account and it is very well possible that $a_p=a_q$ if $p < q$.
The second solution has a quantifier error: given $q$ there is not necessarily one singe $x$ such that $f_r(x)\neq f_q(x)$ for all $r > q$; the $x$ will vary with $r$.
The following is an example defined on the interval $[0,1]$. Consider the standard Cantor set, $C$, in the unit interval $[0,1]$ and let $f:[0,1]\to[0,1]$ be the function, defined by $f(x)=d(x,C)$. For $t\in C$ define $f_t$ by $f_t(x)=f(x)$ if $x\le t$ and $f_t(x)=0$ if $x\ge t$. Then $t < s$ implies $f_t < f_s$.
There is not. Suppose that $\langle P,\le\rangle$ is such a partial order, with associated continuous functions $f_p:[0,1]\to\mathbb{C}$ for $p\in P$.
For $p\in P\,$ let $S(p)=\{x\in[0,1]:f_p(x)\ne 0\}$, and let $K(p)=\operatorname{cl}S(p)$; if $p,q\in P\,$ with $p<q$, the hypotheses imply that $f_q\upharpoonright K(p)=f_p\upharpoonright K(p)$ and hence that $K(q)\supseteq K(p)$. Clearly $f_q=f_p$ if $K(q)=K(p)$, so $K(q)\supsetneqq K(p)$ whenever $p,q\in P$ with $p<q$. (I am assuming that the functions are distinct; otherwise the answer is trivially yes.)
For each $p\in P\,$ let $I(p)$ be an order-component of $S(p)$; $I(p)$ has one of the forms $[0,a_p)$, $(a_p,b_p)$, and $(a_p,1]$, where in each case $a_p\in K(p)\setminus S(p)$. For each $p\in P\,$ there is a positive rational $r_p$ such that one of the intervals $L(p)=(a_p-r_p,a_p)$ and $R(p)=(a_p,a_p+r_p)$ is a subset of $I(p)$. Since $P\,$ is uncountable, there must be a positive rational $r$ and an uncountable $Q\subseteq P$ such that $r_p=r$ for every $p\in Q$. We may further assume without loss of generality that $L(p)\subseteq I(p)$ for every $p\in Q$.
Suppose that $p,q\in Q$ with $p<q$. Then $a_p\in K(p)$, so $f_q(a_p) = f_p(a_p) = 0$, and $a_p\notin L(q)$. On the other hand, $a_q\in K(q)\setminus S(q)\subseteq K(q)\setminus S(p)$, so $a_q\notin L(p)$. Thus, either $a_p\le a_q-r$, or $a_q\le a_p-r$, and in either case $L(p)\cap L(q)=\varnothing$. But then $\{L(p):p\in Q\}$ is an uncountable family of pairwise disjoint open subsets of $[0,1]$, which is impossible.